Chapter 9.1, Problem 82E

To determine

Torepresent:The sum of the series $\left[1-{\left(\frac{1}{6}\right)}^2\right]+\left[1-{\left(\frac{2}{6}\right)}^2\right]+\mathrm{...}+\left[1-{\left(\frac{6}{6}\right)}^2\right]$ in the form ofsigma notation.

Answer to Problem 82E

The sum of the series $\left[1-{\left(\frac{1}{6}\right)}^2\right]+\left[1-{\left(\frac{2}{6}\right)}^2\right]+\mathrm{...}+\left[1-{\left(\frac{6}{6}\right)}^2\right]$ is ${\displaystyle \sum _{i=1}^6\left[1-{\left(\frac{i}{6}\right)}^2\right]}$ .

Explanation of Solution

Given information:

The given series is $\left[1-{\left(\frac{1}{6}\right)}^2\right]+\left[1-{\left(\frac{2}{6}\right)}^2\right]+\mathrm{...}+\left[1-{\left(\frac{6}{6}\right)}^2\right]$ .

Calculation:

Let’s a sequence has $n$ terms and it can be represented as follows.

  ${\displaystyle \sum _{i=1}^n{a}_i}=a_1+a_2+a_3+a_4+\mathrm{...}+a_n$

Here, the index of summation is $i$ , the lower limit of summation is 1 and the upper limit of summation is $n$ .

In the given sequence, there are 6 terms. So, the sum of the series is,

  $\left[1-{\left(\frac{1}{6}\right)}^2\right]+\left[1-{\left(\frac{2}{6}\right)}^2\right]+\mathrm{...}+\left[1-{\left(\frac{6}{6}\right)}^2\right]={\displaystyle \sum _{i=1}^6\left[1-{\left(\frac{i}{6}\right)}^2\right]}$

Therefore, the sum of the series $\left[1-{\left(\frac{1}{6}\right)}^2\right]+\left[1-{\left(\frac{2}{6}\right)}^2\right]+\mathrm{...}+\left[1-{\left(\frac{6}{6}\right)}^2\right]$ in the form of sigma notation is ${\displaystyle \sum _{i=1}^6\left[1-{\left(\frac{i}{6}\right)}^2\right]}$ .